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Jack D'Aurizio mentioned several sums of the form $\sum_{k=0}^n {2n \choose n+k} (-1)^k k^m $ for non-negative integer $m$ in his answer to my question here.

I quote him here, and I note that each of the sums in the large n limit save for the $m=2$ case looks like $O(1){2n \choose n} $, even though the k-dependence looks quite different:

$$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k = \frac{1}{2}\binom{2n}{n}$$ $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k = -\frac{n}{4n-2}\binom{2n}{n} \approx -\frac{1}{4}\binom{2n}{n}$$ $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^2 = 0 $$ $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^3 = \frac{n^2}{2(2n-1)(2n-3)}\binom{2n}{n} \approx \frac{1}{8} \binom{2n}{n}$$

With help from Mathematica, I see the following large-n limit:

$$\sum_{k=1}^{n}\binom{2n}{n+k}(-1)^k \frac{1}{k} \approx -\ln(2)\binom{2n}{n}$$

Since ${2n \choose n}$ increases with n, we can unify all these sums, modulo some O(1) terms, in the form $\sum_{k=1}^n {2n \choose n+k} (-1)^k k^m $. If there's a closed form for all such sums, that would be amazing, but given the increasingly complicated nature of sums of higher powers of k, I think that looking at the large-n limit with its potentially simple form is more fruitful.

I think these sums are pretty neat! Typically one would expect that a decaying dependence on k would shrink the sum, but for large n the magnitude of this $m=-1$ sum looks larger than that of the $m=0$ case, probably because of the interplay with the $(-1)^k$ term. I anticipate that for larger magnitude, negative powers $m$ that the magnitude would decrease. I'm not confident about larger magnitude, positive powers $m$ especially when considering the funny trends from $m=0$ to $m=3$.

Thus my question is whether for all integer powers m the sum $\sum_{k=1}^n {2n \choose n+k} (-1)^k k^m $ takes the form $f(m) {2n \choose n}$ at large n and what the function f(m) might be.

user196574
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    A useful identity is $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k e^{kt} = \binom{2n}{n}\phantom{}_2 F_1(1,-n;n+1;e^{t}), $$ leading to the fact that the wanted sums can be computed by applying $\frac{d^m}{dt^m}$, then evaluating at $t=0$. – Jack D'Aurizio Jun 19 '18 at 22:19
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    In particular the $O(1)$ part for $m\in{4,5,6,7,8,9,10}$ is $$0,-\frac{1}{4},0,\frac{17}{16},0,-\frac{31}{4},0.$$ By the Lagrange inversion theorem (https://math.stackexchange.com/questions/2741644/requesting-basic-explanation-of-the-lagrange-inversion-theorem/2741758#2741758), these coefficients should be related with the Taylor series of the inverse function of $\frac{\sqrt{z}}{\sin\sqrt{z}}$. – Jack D'Aurizio Jun 19 '18 at 22:32

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It's bizarre that a formula I dug out of my garage from over 20 years ago answers two different problems on the same day. The first formula I put on MSE 2824592 just hours ago.

$$ (A)\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}. $$

For $s=2m,$ it is easily seen that RHS is 0. Set $s=2m+1$ and expand the ratio of gamma functions,

$$ \sum_{k=1}^n (-1)^{k} \binom{2n}{n+k} k^{2m+1} = \binom{2n}{n} (-1)^{m+1}\int_0^\infty \frac{dx \, \,x^{2m+1}}{\sinh{\pi x}} \big(1+\frac{x^2}{n} + ...\big). $$ The integrals are known by Mathematica and the result is $$ \sum_{k=1}^n (-1)^{k} \binom{2n}{n+k} k^{2m+1} = \binom{2n}{n} (-1)^{m+1}\big(c_{2m+2} + \frac{c_{2m+4}}{n} + ... \big) $$ where $$c_{n}= (2-2^{1-n})\frac{\zeta(n)}{\pi^n}(n-1)! $$

user321120
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  • That's likely because I asked both of them! Perhaps you could try the question-and-answer format with your formula? That way you could show your proof and see if others may have a slick way. – user196574 Jun 20 '18 at 20:36
  • Hello user196574. I've answered questions, but not asked. Isn't it bad form to ask a question for which you already have an answer? I could ask, hundreds of times, does anyone have a slick proof of what I think is a new result? – user321120 Jun 21 '18 at 15:51